Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $z = \dfrac{-5x + 5}{2x^2 + 32x + 120} \times \dfrac{x^2 + 15x + 54}{2x + 18} $
Answer: First factor out any common factors. $z = \dfrac{-5(x - 1)}{2(x^2 + 16x + 60)} \times \dfrac{x^2 + 15x + 54}{2(x + 9)} $ Then factor the quadratic expressions. $z = \dfrac {-5(x - 1)} {2(x + 6)(x + 10)} \times \dfrac {(x + 6)(x + 9)} {2(x + 9)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac {-5(x - 1) \times (x + 6)(x + 9) } { 2(x + 6)(x + 10) \times 2(x + 9)} $ $z = \dfrac {-5(x + 6)(x + 9)(x - 1)} {4(x + 6)(x + 10)(x + 9)} $ Notice that $(x + 6)$ and $(x + 9)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac {-5\cancel{(x + 6)}(x + 9)(x - 1)} {4\cancel{(x + 6)}(x + 10)(x + 9)} $ We are dividing by $x + 6$ , so $x + 6 \neq 0$ Therefore, $x \neq -6$ $z = \dfrac {-5\cancel{(x + 6)}\cancel{(x + 9)}(x - 1)} {4\cancel{(x + 6)}(x + 10)\cancel{(x + 9)}} $ We are dividing by $x + 9$ , so $x + 9 \neq 0$ Therefore, $x \neq -9$ $z = \dfrac {-5(x - 1)} {4(x + 10)} $ $ z = \dfrac{-5(x - 1)}{4(x + 10)}; x \neq -6; x \neq -9 $